It is shown in section 6.1 that for the E<U₀ potential step, $\mathbf{B}\mathbf{=}\mathbf{-}\frac{\mathbf{\alpha }\mathbf{+}\mathbf{ik}}{\mathbf{\alpha }\mathbf{-}\mathbf{ik}}\mathbf{A}$. Use it to calculate the probability density to the left of the step:

${\left|{\psi }_{x<0}\right|}^{\mathbf{2}}\mathbf{=}{\left|{\mathrm{Ae}}^{\mathrm{ikx}}+{\mathrm{Be}}^{-\mathrm{ikx}}\right|}^{\mathbf{2}}$

1. Show that the result is, $\mathbf{4}{\left|A\right|}^{\mathbf{2}}\mathbf{}{\mathbf{sin}}^{\mathbf{2}}\left(\mathrm{kx}-\theta \right)$where $\mathbf{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(k/\alpha \right)$. Because the reflected wave is of the same amplitude as the incident, this is a typical standing wave pattern varying between 0and $\mathbf{4}\mathbf{A}\mathbf{*}\mathbf{A}$.
2. Determine data-custom-editor="chemistry" $\mathbf{\theta }$and Din the limits kanddata-custom-editor="chemistry" $\mathrm{\alpha }$tend to 0and interpret your results.

Probability density is a function whose value at any given sample in the sample space gives the likelihood of that random value would be close to that sample.

Consider the given function

${\left|{\mathrm{\psi }}_{\mathrm{x}<0}\right|}^2={\left|{\mathrm{Ae}}^{\mathrm{ikx}}+{\mathrm{Be}}^{-\mathrm{ikx}}\right|}^2$

$\mathrm{B}=-\frac{\mathrm{\alpha }+\mathrm{ik}}{\mathrm{\alpha }-\mathrm{ik}}\mathrm{A}$ ….. (1)

Here, data-custom-editor="chemistry" $\mathrm{\psi }$is the wave function, data-custom-editor="chemistry" $\mathrm{A},\mathrm{B},\mathrm{\alpha }$is the Arbitrary constants$\mathrm{k}=\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}$

${\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|{\mathrm{Ae}}^{\mathrm{ikx}}+{\mathrm{Be}}^{-\mathrm{ikx}}\right|}^2$

$\mathrm{B}=-\frac{\mathrm{\alpha }+\mathrm{ik}}{\mathrm{\alpha }-\mathrm{ik}}\mathrm{A}$ …..(2)

After using equation (1) in equation (2), and taking common from the Right-Hand-Side, you get,

role="math" localid="1660032485327" ${\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|\mathrm{A}\right|}^2{\left|{\mathrm{e}}^{\mathrm{ikx}}-\frac{\mathrm{\alpha }+\mathrm{ik}}{\mathrm{\alpha }-\mathrm{ik}}{\mathrm{e}}^{-\mathrm{ikx}}\right|}^2$

After splitting the squared expression and further solving it as:

$$\begin{gathered} {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|\mathrm{A}\right|}^2\left({\mathrm{e}}^{\mathrm{ikx}}-\frac{\mathrm{\alpha }+\mathrm{ik}}{\mathrm{\alpha }-\mathrm{ik}}{\mathrm{e}}^{-\mathrm{ikx}}\right)\left({\mathrm{e}}^{-\mathrm{ikx}}-\frac{\mathrm{\alpha }+\mathrm{ik}}{\mathrm{\alpha }-\mathrm{ik}}{\mathrm{e}}^{\mathrm{ikx}}\right) \\ {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|\mathrm{A}\right|}^2\left(1-\frac{\mathrm{\alpha }+\mathrm{ik}}{\mathrm{\alpha }-\mathrm{ik}}{\mathrm{e}}^{-2\mathrm{ikx}}-\frac{\mathrm{\alpha }+\mathrm{ik}}{\mathrm{\alpha }-\mathrm{ik}}{\mathrm{e}}^{2\mathrm{ikx}}+1\right) \end{gathered}$$

Using${\mathrm{\alpha }}^2+{\mathrm{k}}^2$as the common denominator solve as:

$$\begin{gathered} {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|\mathrm{A}\right|}^2\left(\frac{2\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)-{\left(\mathrm{\alpha }+\mathrm{ik}\right)}^2{\mathrm{e}}^{-2\mathrm{ikx}}-{\left(\mathrm{\alpha }-\mathrm{ik}\right)}^2{\mathrm{e}}^{2\mathrm{ikx}}}{\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)}\right) \\ {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|\mathrm{A}\right|}^2\left(\frac{2\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)-\left({\mathrm{\alpha }}^2-{\mathrm{k}}^2+2\mathrm{ik\alpha }\right){\mathrm{e}}^{-2\mathrm{ikx}}-\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2-2\mathrm{ik\alpha }\right){\mathrm{e}}^{2\mathrm{ikx}}}{\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)}\right) \\ {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|\mathrm{A}\right|}^2\left(\frac{2\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)-\left({\mathrm{\alpha }}^2-{\mathrm{k}}^2\right)\left({\mathrm{e}}^{-2\mathrm{ikx}}+{\mathrm{e}}^{2\mathrm{ikx}}\right)+2\mathrm{ik\alpha }\left({\mathrm{e}}^{-2\mathrm{ikx}}-{\mathrm{e}}^{2\mathrm{ikx}}\right)}{\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)}\right) \end{gathered}$$

Now, after using Euler’s equation:${\mathbf{e}}^{\mathbf{i\theta }}\mathbf{=}\mathbf{cos\theta }\mathbf{+}\mathbf{isin\theta }$ and further solving it as:

$$\begin{gathered} {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|\mathrm{A}\right|}^2\left(\frac{2\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)-\left({\mathrm{\alpha }}^2-{\mathrm{k}}^2\right)2\cos \left(2\mathrm{kx}\right)+4\mathrm{k\alpha sin}\left(2\mathrm{kx}\right)}{\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)}\right) \\ {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2={\left|\mathrm{A}\right|}^2\left(\frac{2\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)-\left({\mathrm{\alpha }}^2-{\mathrm{k}}^2\right)\left({\cos }^2\mathrm{kx}-{\sin }^2\mathrm{kx}\right)+4\mathrm{k\alpha }\left(2\mathrm{sinkxcoskx}\right)}{\left({\mathrm{\alpha }}^2+{\mathrm{k}}^2\right)}\right) \\ {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2=4{\left|\mathrm{A}\right|}^2{\left(\frac{\mathrm{\alpha }}{\sqrt{{\mathrm{\alpha }}^2+{\mathrm{k}}^2}}\mathrm{sinkx}-\frac{\mathrm{k}}{\sqrt{{\mathrm{\alpha }}^2+{\mathrm{k}}^2}}\mathrm{coskx}\right)}^2 \\ {\left|{\mathrm{\psi }}_{\mathrm{I}}\right|}^2=4{\left|\mathrm{A}\right|}^2{\left(\mathrm{cos\theta sinkx}-\mathrm{sin\theta coskx}\right)}^2 \end{gathered}$$

Hence, the wave function is obtained as ${\left|\mathrm{\psi }\right|}^2=4{\left|\mathrm{A}\right|}^2{\sin }^2\left(\mathrm{kx}-\mathrm{\theta }\right)$

Here,$\mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{k}}{\mathrm{\alpha }}\right)$

If k=0, $\mathrm{\theta }=0^{\mathrm{o}}$and D=0. The wave is zero at the step. The step is relatively so high that the wave doesn’t penetrate it. If $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{=}{\mathbf{90}}^{\mathbf{o}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{D}\mathbf{=}\mathbf{2}\mathbf{A}$. The wave is maximum at the step and there is much penetration.

Hence, the required parameters are obtained as $\mathbf{\theta }\mathbf{=}{\mathbf{90}}^{\mathbf{o}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{D}\mathbf{=}\mathbf{2}\mathbf{A}$.